Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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In a commutative ring, (S)(S) consists of finite sums ∑risi\sum r_i s_i, and (a)=Ra(a)=Ra

Statement

In a commutative ring, (S)(S) consists of finite sums ∑risi\sum r_i s_i, and (a)=Ra(a)=Ra.

The empty sum is included and equals 00.

Facts & Assumptions

Given: A commutative ring RR and a subset S⊆RS\subseteq R.

[L1]

(S)(S) is the intersection of ideals containing SS (The ideal generated by a subset and principal ideals).

[L2]

The ideal criterion uses subtraction and absorption (Ideal criteria and intersections of ideals).

[L3]

Multiplication in a commutative ring commutes (Commutative ring).

Proof

technique · direct
1.1

Let JJ be the finite sums ∑risi\sum r_is_i; it contains SS and is closed under subtraction and multiplication by arbitrary ring elements.

L1L2L3givenalgebra
2.1

Thus JJ is an ideal containing SS, while every ideal containing SS contains each such finite sum.

step 1.1L1L2L3givenalgebra
3.1

Hence J=(S)J=(S); taking S={a}S=\{a\} gives (a)=Ra(a)=Ra.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 18 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources